1. www.metoffice.gov.uk © Crown Copyright 2017, Met Office
Where do the horizontal waves go? Propagation, reflection and absorption

2. Propagation, reflection and absorption
wavenumber k = 1
Maps of l2
wavenumber k = 2
Orange and red: propagation allowed
Dark blue regions are easterly winds: absorption of all waves
Light blue regions, strong winds: reflection of short (high wavenumber) waves
Absorptive
Reflective
wavenumber k = 3
wavenumber k = 4
© Crown Copyright 2019, Met Office

3. Barotropic Rossby Waves
Simplest case: barotropic, no forcing → constant absolute vorticity
𝐷 𝜁+𝑓 𝐷𝑡 =0
𝛽= 𝜕𝑓 𝜕𝑦
Beta plane: Coriolis parameter f is linear
𝜕 𝜕𝑡 +𝑢 𝜕 𝜕𝑥 +𝑣 𝜕 𝜕𝑦 𝜁+𝛽𝑣=0
We’ve got rid of the forcing terms on the RHS of our original equation

4. Barotropic Rossby Waves (cont.)
𝜕 𝜕𝑡 +𝑢 𝜕 𝜕𝑥 +𝑣 𝜕 𝜕𝑦 𝜁+𝛽𝑣=0
From previous slide
Perturbation to zonal mean flow:
𝑢= 𝑢 + 𝑢 ′
v= 𝑣 ′
𝜁= 𝜁 + 𝜁 ′
By construction and hence and similarly for other quantities
𝑢′ =0
𝑢 𝑢′ = 0
𝑢 ′ =− 𝜕 𝜓 ′ 𝜕𝑦
𝑣 ′ = 𝜕 𝜓 ′ 𝜕𝑥
Perturbation streamfunction:
𝜁 ′ = 𝛻 2 𝜓 ′
𝜕 𝜕𝑡 + 𝑢 𝜕 𝜕𝑥 𝛻 2 𝜓 ′ + 𝛽− 𝑢 𝑦𝑦 𝜕 𝜓 ′ 𝜕𝑥 =0
Split flow into mean and perturbations
Assume mean flow is only in zonal direction
Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

5. Barotropic Rossby Waves (cont.)
𝜕 𝜕𝑡 + 𝑢 𝜕 𝜕𝑥 𝛻 2 𝜓 ′ + 𝛽− 𝑢 𝑦𝑦 𝜕 𝜓 ′ 𝜕𝑥 =0
From previous slide
Try a wave solution:
𝜓 ′ =𝐴 𝑒 𝑖 𝑘𝑥+𝑙𝑦−𝜔𝑡
k = zonal wavenumber = 2π/Lx where Lx is the zonal wavelength
l = meridional wavenumber = 2π/Ly
ω = frequency = 2π/T where T is the period
A solution exists only if:
𝜔= 𝑢 𝑘− 𝛽− 𝑢 𝑦𝑦 𝑘 𝑘 2 + 𝑙 2
Rossby wave dispersion relation
Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

6. Rossby Wave Dispersion
𝜔= 𝑢 𝑘− 𝛽− 𝑢 𝑦𝑦 𝑘 𝑘 2 + 𝑙 2
Rossby wave dispersion relation
(From previous slide)
Consider a 1D wave propagating in x direction, phase ϕ(x,t)=kx-ωt
The phase is constant for an observer moving at the phase speed
𝐷𝜙 𝐷𝑡 = 𝐷 𝐷𝑡 𝑘𝑥−𝜔𝑡 =𝑘 𝐷𝑥 𝐷𝑡 −𝜔=0
𝑐 𝑥 = 𝐷𝑥 𝐷𝑡 = 𝜔 𝑘
Hence phase speed
So dispersion relation gives the phase speed for different wave numbers
Wavenumbers move at different speeds so wave packet will spread out (disperse)
cx generally westward relative to mean flow
Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

7. Group Velocity 𝑐 𝑔𝑦 = 𝜕𝜔 𝜕𝑙 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 𝛽− 𝑢 𝑦𝑦
𝑐 𝑔𝑦 = 𝜕𝜔 𝜕𝑙 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 𝛽− 𝑢 𝑦𝑦
𝑐 𝑔𝑥 = 𝜕𝜔 𝜕𝑘 = 2 𝑢 2 𝑘 2 𝛽− 𝑢 𝑦𝑦
Eastward propagation faster for shorter wavelengths (higher k)
Meridional propagation possible only if term in {} is positive
stops in strong winds (reflections)
stops at zero wind line (absorption)
easier for longer wavelengths (low k)
can in principle be prevented by curvature of wind field (uyy)
Square root has 2 values → N and S propagation (symmetry for equatorial sources)
Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics

8. Rossby wave absorption and reflection
Meridional wavenumber
(from dispersion relation with ω=0)
𝑙= 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘
Maps of l2
wavenumber k = 1
wavenumber k = 2
Orange and red: propagation allowed
Dark blue regions are easterly winds: absorption of all waves
Light blue regions, strong winds: reflection of short (high wavenumber) waves
Orange and red: propagation allowed
Dark blue regions are easterly winds: absorption of all waves
Light blue regions, strong winds: reflection of short (high wavenumber) waves
Absorptive
wavenumber k = 3
Reflective
wavenumber k = 4
Scaife et al 2016

9. Ray Tracing 𝑐 𝑔𝑦 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 2 1 2 𝛽− 𝑢 𝑦𝑦
𝑐 𝑔𝑦 = 2 𝑢 2 𝑘 𝛽− 𝑢 𝑦𝑦 𝑢 − 𝑘 𝛽− 𝑢 𝑦𝑦
𝑐 𝑔𝑥 = 2 𝑢 2 𝑘 2 𝛽− 𝑢 𝑦𝑦
Group velocity
𝛽= 2Ω cos Φ 𝑎
k= 2𝜋 𝐿 𝑥
𝑢 =
Calculate cgx and cgy
Calculate new position after timestep (2 hours)
Repeat
Holton, Introduction to dynamic meteorology ; James, Introduction to circulating atmospheres ; Vallis, Atmospheric and oceanic fluid dynamics ; Scaife et al 2016

10. Ray Tracing From the W Pacific: Wave2 propagates Wave3 reflects
propagation (k = 2)
propagation (k = 3)
propagation k = 4
Rossby Wave Ray Paths
From the W Pacific:
Wave2 propagates
Wave3 reflects
From the E Pacific:
Wave4 reflects
k = 2
k = 2
k = 3
k = 4
Scaife et al 2016

11. Rossby Wave Sources 𝑆=− 𝜁𝛻. 𝒗 𝜒 + 𝒗 𝜒 .𝛻𝜁 =−𝛻. 𝒗 𝜒 𝜁
Decompose velocity in rotational (vψ) and divergent(vχ) parts
Rossby wave source
𝑆=− 𝜁𝛻. 𝒗 𝜒 + 𝒗 𝜒 .𝛻𝜁 =−𝛻. 𝒗 𝜒 𝜁
Stretching Advection
Rossby waves are generated where divergent flow interacts with vorticity gradient
Sardeshmukh and Hoskins ; James, Introduction to circulating atmospheres ; Scaife et al 2016

12. Ray Tracing: teleconnections
TEP TA
Rays intersect main centres (of 200 hPa geopotential height anomalies) from a few common sources
Wave 2, 3 mainly responsible as wave 4 rarely propagates
We have a theory for the teleconnections from tropical rainfall IO
TWP
Scaife et al 2016, Zhao et al “Dynamics of an Interhemispheric Teleconnection across the Critical Latitude through a Southerly Duct during Boreal Winter”, J Clim 2015

13. Vertically propagating Rossby waves
© Crown Copyright 2017, Met Office

14. Observed zonal mean flow
July
Heavy contours = westerlies
Thin contours = easterlies
Coloured contours = isotherms
Stratospheric night jet
Tropospheric westerly jet
From surface to jet core, u gets stronger with height as theta increases – the thermal wind relation…
North pole
South pole

15. Equations of motion (hydrostatic approx.)
𝐷𝑢 𝐷𝑡 −𝑓𝑣+ 𝜕Φ 𝜕𝑥 =𝑋
𝐷𝑣 𝐷𝑡 −𝑓𝑢+ 𝜕Φ 𝜕𝑦 =𝑌
Horizontal momentum equations on mid-latitude beta plane
𝜕Φ 𝜕𝑧 = 𝑅 𝐻 𝑇
𝜕𝑢 𝜕𝑥 + 𝜕𝑣 𝜕𝑦 + 1 𝜌 0 𝜕( 𝜌 0 𝑤) 𝜕𝑧 =0
𝐷𝑇 𝐷𝑡 + 𝜅𝑇 𝐻 𝑤= 𝐽 𝑐 𝑝
Vertical momentum equation (pressure coordinates) in hydrostatic approximation
𝑧=𝐻 ln 𝑝 𝑅 𝑝
Continuity equation (pressure coordinates)
Thermodynamic energy equation
(Following Holton ch. 10)

16. Eulerian averaging
We can expand any variable in terms of its longitudinally averaged value and a perturbation
We can continue our crossing out spree using the typical magnitudes of the various variables (and hence their products):
𝐴= 𝐴 +𝐴′
We substitute into the equations of motion, and zonally average both sides.
Then we go on a crossing out spree using: 𝐴
𝐴′ =0
𝐴 𝐴′ =0
is independent of x
(and of course the usual crossing out of higher order terms) so
(also applies to terms like )
𝐴 𝐵′ =0

17. After a lot of expanding and crossing out…
Starting from the equation for u
Starting from the equation for v
Zonal mean zonal momentum equation
𝜕 𝑢 𝜕𝑡 − 𝑓 0 𝑣 = − 𝜕 𝑢 ′ 𝑣 ′ 𝜕𝑦 + 𝑋
𝜕 𝑇 𝜕𝑡 + 𝑁 2 𝐻 𝑅 𝑤 = − 𝜕 𝑣 ′ 𝑇′ 𝜕𝑦 + 𝐽 𝑐 𝑝
Geostrophic balance
𝑓 0 𝑢 = − 𝜕 Φ 𝜕𝑦
𝑓 0 𝜕 𝑢 𝜕𝑧 + 𝑅 𝐻 𝜕 𝑇 𝜕𝑦 = 0
Eq 1
Eq 3
Zonal mean thermodynamic energy equation
And combining with the hydrostatic approx
Thermal wind equation
Eq 2
Eq 4
with buoyancy frequency
Eq 4 : as T decreases with height, u increases
Eqs 1 and 2 : unless we have some sort of meridional circulation, then the eddy heat flux divergence and eddy momentum flux divergence would destroy the thermal wind balance (it doesn’t destroy it, but it does reduce it).
𝑁 2 ≡ 𝑅 𝐻 𝜅 𝑇 0 𝐻 − 𝑑 𝑇 0 𝑑𝑧
Remember this when we get to Rossby waves

18. Schematic of the meridional circulation
Reduced thermal wind
Hadley cell
𝑓 𝑣 <0
Ferrel cell
𝜕 𝑢′𝑣′ 𝜕𝑦 ≪0
Latent heating
Radiative cooling
Adiabatic warming
Observed max heat flux
Adiabatic cooling
𝑣′𝑇′ 𝑚𝑎𝑥
𝜕 𝑣′𝑇′ 𝜕𝑦 >0
𝜕 𝑣′𝑇′ 𝜕𝑦 <0
about 50 N
𝜕 𝑢′𝑣′ 𝜕𝑦 <0
𝑓 𝑣 >0
Equator
High latitudes

19. Take-home messages from this section
Given that we have eddy heat and momentum fluxes, in the absence of a meridional circulation the zonal winds wouldn’t stay in geostrophic balance
The eddy momentum flux divergence
and eddy heat flux divergence
together drive a meridional circulation which restores geostrophic balance
The two working together is what sets the background for EP fluxes
Hence existence of thermally indirect Ferrel cell
But also has the effect of reducing the thermal wind (shear is less pronounced with height up to jet maximum than we’d otherwise expect)
However - 𝑢 does still increase with height – relevant in context of wave breaking
𝜕 𝑢′𝑣′ 𝜕𝑦
𝜕 𝑣 ′ 𝑇′ 𝜕𝑦

20. Setting up the problem
Reminder: still working in log-pressure vertical coordinates
Scale height
Assume isothermal atmosphere with density
and Brunt-Väisälä frequency
𝑧=𝐻 ln 𝑝 𝑅 𝑝
𝐻=𝑅 𝑇 0 𝑔
𝜌 𝑅 𝑁
(Explanation lifted from James, 1994
NB my slides make minor notational changes for consistency with Holton.)
Density varies with height
𝜌 𝑅 = 𝜌 0 𝑒 − 𝑧 𝐻
Basic state consists of constant zonal wind 𝑈
Partition stream function into zonal mean and eddy parts
𝜓= −𝑈𝑦+ 𝜓 ′
Mid latitude beta plane

21. Wave-like solutions
Using we linearize the potential vorticity equation
𝜓= −𝑈𝑦+ 𝜓 ′
𝑈 𝜕 𝜕𝑥 𝛻 2 ψ ′ + 𝑓 2 𝜌 𝑅 𝜕 𝜕𝑧 𝜌 𝑅 𝑁 2 𝜕𝜓′ 𝜕𝑧 𝛽 𝜕𝜓 𝜕𝑥 = 0
Decrease in density with height
Want to keep wave activity the same
Hence increase in amplitude with height
and look for wavelike solutions
𝜓 ′ = Ψ 𝑧 𝑒 𝑧 2𝐻 𝑒 𝑖 𝑘𝑥+𝑙𝑦
Substituting this solution back in, we get:
𝑑 2 Ψ 𝑑𝑧 𝑁 2 𝑓 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Ψ = 0
James 1994
Eq 5

22. Wavelike solutions – conditions for vertical propagation
Needs to be positive for wavelike solutions
𝑑 2 Ψ 𝑑𝑧 𝑁 2 𝑓 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Ψ = 0
Easterly – no waves!
Moderate westerlies,
Low total wave number
Strong westerly – no waves!
Large total wave number
– no waves!
𝐾 = 𝑘 2 + 𝑙 2
James 1994

23. Wave breaking as height increases
We want wavelike solutions of
𝑑 2 Ψ 𝑑𝑧 𝑁 2 𝑓 𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2 Ψ = 0
Eq 5
But we know u increases with height
𝑓 0 𝜕 𝑢 𝜕𝑧 + 𝑅 𝐻 𝜕 𝑇 𝜕𝑦 = 0
Eq 4

24. Wave breaking as height increases
𝛽 𝑈 − 𝐾 2 − 𝑓 2 4 𝑁 2 𝐻 2
Wave of given wavenumber propagates vertically
In vertical
U increases with height
We reach a height at which
Wave crosses critical line from propagating to evanescent solutions
𝛽 𝑈 < 𝐾 2
James 1994

25. Rossby Wave breaking and sudden stratospheric warmings
Shift in, and disruption of, polar night jet over 4 days
Arrows are Eliassen-Palm vectors
© Crown Copyright 2019, Met Office
Coy, JAS, 2009, “Planetary wave breaking… in SSW of Jan 2006”

26. Rossby Wave breaking and sudden stratospheric warmings
Example of this in the next lecture, from last winter.
Easterly anomalies in stratosphere gradually propagate downwards
© Crown Copyright 2019, Met Office

27. www.metoffice.gov.uk © Crown Copyright 2017, Met Office
Summary

28. Summary Horizontally propagating Rossby Waves
In terms of vorticity – physical explanation
The vorticity equation
Wave solutions and the dispersion relation
Propagation, absorption and reflection, and ray-tracing
Vertically propagating Rossby Waves
Thermal wind – stratospheric night jet
The meridional circulation
Wave breaking with increasing height
Wave propagation and absorption helps explain teleconnections between the tropics and extra-tropics
Wave breaking in the stratosphere explains how disruption of the stratospheric polar vortex impacts the surface
We can successfully represent these in dynamical models
This underlies a lot of the skill in dynamical models
© Crown Copyright 2019, Met Office

29. www.metoffice.gov.uk © Crown Copyright 2017, Met Office
Questions
© Crown Copyright 2017, Met Office